Even though the OP does not seem to be active in this tread any longer, I'm going to provide an answer anyway, since it seems like such an interesting question.
First of all, even if a closed system undergoes a process in which it is in contact with a constant pressure surroundings throughout the entire process, and this pressure is the same as that of the system in its initial and final states, the heat absorbed Q is still not necessarily equal to the change in enthalpy unless PV work is the only form of work being done. For example, if there is a stirrer doing work to agitate the system (and cause irreversible viscous heating), and no PV work occurs, and the system is in contact with a constant temperature bath throughout, the change in enthalpy will be equal to zero, but Q will be equal to minus the amount of work that the stirrer does on the system.
Secondly, assume that throughout the irreversible process, the system undergoes a change in which it is in contact with a single constant temperature reservoir at the same temperature as the initial and final temperatures of the system, and is also in contact with a constant pressure surroundings at the same pressure as that of the system in its initial and final states. But consider that a reversible process between the same initial and final end states does not have to resemble the irreversible path in any way whatsoever, as long as it matches the initial and final temperatures and pressures (and species concentrations if chemical reaction is involved). Even if the boundary temperature for the reversible path is held constant at the same value as the irreversible path, the pressure of the surroundings during the reversible path definitely does not have to be (and will not be) constant throughout the reversible path. And, given that this is the case, the heat Q for the reversible path would not be required to be the same as that for the irreversible path, even if the enthalpy changes for both paths is the same. In virtually all cases, it will actually not be possible to devise a reversible path at a constant surrounding pressure between the same pair of end states. So the heat Q for the reversible path will not be equal to the change in enthalpy for the reversible path (and thus the change in enthalpy for the irreversible path).